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pychu [463]
3 years ago
5

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of

her shadow changing when she is 15 ft from the base of the light?
Physics
2 answers:
Soloha48 [4]3 years ago
7 0

Answer:

The length of her shadow is changing at the rate  -2 m/s

Explanation:

Let the height oh the street light, h = 22 ft

Let the height of the woman, w = 5.5 ft

Horizontal distance to the street light = l

length of shadow = x

h/w = (l + x)/x

22/5.5 =  (l + x)/x

4x = l + x

3x = l

x = 1/3 l

taking the derivative with respect to t of both sides

dx/dt = 1/3 dl/dt

dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

dx/dt = 1/3 * (-6)

dx/dt = -2 m/s

sveticcg [70]3 years ago
5 0

Answer:

Explanation:

By applying similar triangles rule,

22/5.5 = (x + s)/s

s = 5.5/22 × (x + s)

ds/dt = 5.5/22 × (dx/dt + ds/dt)

16.5/22 × ds/dt = 5.5/22 × dx/dt

16.5/22 × ds/dt = 5.5/22 × -6

ds/st = -2 ft/s.

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When the ball goes up it’s potential energy

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Which two pieces of evidence most directky support the idea that all matter in the universe began as a sibgke point of energy.
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YOUR ANSWER IS GIVEN BELOW:

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The abundance of helium in the universe and the redshift of light from the galaxies.

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The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver
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Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

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Stage 2:

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Stage 3:

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Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100

Lets find the speed at stage 1

Speed =  \frac{Distance }{Time}

Speed =  \frac{80}{1}

Speed 1= 80 km/hr

The speed at stage 2

Speed =  \frac{Distance }{Time}

Speed =  \frac{200}{2}

Speed 2  = 100 km/hr

The speed at stage 3

Speed =  \frac{Distance }{Time}

Speed =  \frac{x}{4}

Speed 3  = \frac{x}{4}

we kow that average is ,

\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100

\frac{ 80 + 100+ \frac{x}{4} }{3} = 100

\frac{ 180 + \frac{x}{4} }{3} = 100

\frac{ \frac{720 +x}{4} }{3} = 100

\frac{720 +x}{4} \times \frac{1}{3} = 100

\frac{720 +x}{12} = 100

720 +x = 100 \times 12

720 +x = 1200

x = 1200- 720

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Answer:

<h2>A. 6pF</h2>

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4}  = 1.205-1.033\\\\\frac{1}{C_4}  = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\

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<h3>What is a paramagnetic substance?</h3>

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A magnetic field is defined as the field that exists around a magnet that produces a field of force.

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These substances are known as paramagnetic substances because they possess a high number of unpaired electrons.

Other properties of a paramagnetic substance include the following:

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